Students must start practicing the questions from CBSE Sample Papers for Class 10 Maths Basic with Solutions Set 5 are designed as per the revised syllabus.

## CBSE Sample Papers for Class 10 Maths Basic Set 5 with Solutions

Time : 3 hrs

Max. Marks : 80

General Instructions:

- This Question Paper has 5 Sections A, B, C, D and E.
- Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
- Section B has 5 Short Answer-I (SA-I) type questions carrying 2 marks each.
- Section C has 6 Short Answer-II (SA-II) type questions carrying 3 marks each.
- Section D has 4 Long Answer (LA) type questions carrying 5 marks each.
- Section E has 3 Case Based integrated units of assessment questions (4 marks each).

All Questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks and 2 Questions of 5 marks has been provided. An internal choice has been provided in the 2 marks question of Section E. - Draw neat figures wherever required. Take π = 22/7 wherever required if not stated.

Section – A

Section A consists of 20 questions of 1 mark each

Question 1.

The HCF of 85 and 152 is

(a) 1

(b) 17

(c) 13

(d) 19

Answer:

(a) 1

Let a = 85

and b = 152

∴ a = 1 × 5 × 17

and b = 1 × 23 × 19

HCF (85, 152) = 1

Question 2.

The sum of the exponents of the prime factors in the prime factorisation of 198 is

(a) 2

(b) 4

(c) 6

(d) 8

Answer:

(b) 4

Let a = 198

∴ a = 2 × 32 × 11

Required sum = 1 + 2 + 1 = 4

Question 3.

If the point (x, y) is equidistant from the points (2, 1) and (1, – 2), then

(a) x + 3y = 0

(b) 3x + y = 0

(c) x + 2y = 0

(d) 3x + 2y = 0

Answer:

(a) x + 3y = 0

Let the points be P(x, y), 4(2, 1) and B(1, – 2).

Since, P is equidistant from A and B.

∴ AP = BP

⇒ AP^{2} = BP^{2}

⇒ (x – 2)^{2} + (y – 1)^{2} = (x – 1)^{2} + (y + 2)^{2}

[by distance formula]

⇒ x^{2} – 4x + 4 + y^{2} – 2y + 1 = x^{2} – 2x + 1 + y^{2} + 4y + 4

⇒ 2x + 6y = 0

⇒ x + 3y = 0 [dividing by 2]

Question 4.

A die is thrown once the probability of getting an odd number is

(a) \(\frac{1}{3}\)

(b) \(\frac{1}{4}\)

(c) \(\frac{1}{6}\)

(d) \(\frac{1}{2}\)

Answer:

(d) \(\frac{1}{2}\)

When a die is thrown once, we get 1, 2, 3, 4, 5,6

Out of which there are 1, 3, 5 odd numbers.

∴ Required probability = \(\frac{3}{6}=\frac{1}{2}\)

Question 5.

In a ∆ABC, it is given that AB = 6 cm, AC = 8 cm and AD is the bisector of A. Then, BD : DC is

(a) 3 : 4

(b) 9 : 16

(c) 4 : 3

(d) √3 : 2

Answer:

(a) 3 : 4

Given, AD is the bisector of ∠A.

By angle bisector theorem, bisector of an angle divides the opposite side in the same ratio as of two remaining sides.

∴ \(\frac{B D}{D C}=\frac{A B}{A C}\)

⇒ \(\frac{B D}{D C}=\frac{6}{8}\)

⇒ \(\frac{B D}{D C}=\frac{3}{4}\)

Question 6.

If the length of the shadow of a building is decreasing, then the angle of elevation is

(a) decreasing

(b) increasing

(c) remains same

(d) None of these

Answer:

(b) increasing

Here, AB is the building and BE is the initial length of the shadow.

As, the shadow reaches from E to C the angle of elevation increases from 30° to 60°.

Question 7.

The ratio in which Y-axis divides the line segment joining the points P (- 2, 5) and Q (2, – 7) is

(a) 1

(b) \(\frac{1}{2}\)

(c) \(\frac{3}{2}\)

(d) \(\frac{4}{3}\)

Answer:

(a) 1

Suppose Y-axis divides PQ in the ratio λ : 1 at point R,

then the coordinates of R are \(\left(\frac{2 \lambda-2}{\lambda+1}, \frac{-7 \lambda+5}{\lambda+1}\right)\)

Since, R lies on Y-axis

∴ \(\frac{2 \lambda-2}{\lambda+1}\) = 0

⇒ 2λ – 2 = 0

⇒ λ = 1

Question 8.

The expression for the common difference of an AP whose first term is a and nth term is b, is

(a) \(\frac{a-b}{n}\)

(b) \(\frac{b-a}{n}\)

(c) \(\frac{b}{a}\)

(d) \(\frac{b-a}{n-1}\)

Answer:

(d) \(\frac{b-a}{n-1}\)

Let the common difference bed.

We know that a_{n} = a + (n – 1) d

According to the question.

⇒ b = a + (n – 1) d

⇒ d = \(\frac{b-a}{n-1}\)

Question 9.

In ∆DEW, AB || EW. If AD = 2 cm, DE = 6 cm and DW = 15 cm, then the value of DB is

(a) 2 cm

(b) 5 cm

(c) 9 cm

(d) 13 cm

Answer:

(b) 5 cm

⇒ 15 – DB = 2 DB

⇒ 3 DB = 15

⇒ DB = \(\frac{15}{3}\) = 5 cm

Question 10.

AB is a chord of the circle and AOC is its diameter such that ∠BAC = 60°. If CP is the tangent to the circle at the point C, then ∠BCP is equal to

(a) 50°

(b) 60°

(c) 70°

(d) 80°

Answer:

(b) 60°

∠ABC = 60°

[∵ angle in a semi-circle is right angle]

∠A + ∠B + ∠C = 180°

⇒ 60° + 90° + ∠C = 180°

⇒ ∠C = 180° – 150° = 30°

Now, ∠OCP = 90°

[∵ radius is perpendicular to the tangent at the point of contact]

∴ ∠BCP = 90° – ∠OCB

= 90° – 30° = 60°

Question 11.

If sec θ – tan θ = x, then the value of sec θ + tan θ in term of x is

(a) \(\frac{1}{x}\)

(b) x^{2}

(c) \(\frac{x-1}{x}\)

(d) x

Answer:

(a) \(\frac{1}{x}\)

We know that

sec^{2} θ – tan θ = 1

⇒ (sec θ – tan θ) (sec θ + tan θ) = 1

⇒ x (sec θ + tan θ) = 1

⇒ sec θ + tan θ = \(\frac{1}{x}\)

Question 12.

If the system of equations 5x + 2 y = k and 10x + 4y = 3 has infinitely many solutions, then the value of k is

(a) \(\frac{3}{2}\)

(b) \(\frac{1}{2}\)

(c) \(\frac{1}{3}\)

(d) 2

Answer:

(a) \(\frac{3}{2}\)

Given, system of equations is

5x + 2y = k

and 10x + 4y =3

Here a_{1} =5, b_{1} = 2, c_{1} = – k

and a_{2} = 10, b_{2} = 4, c_{2} = – 3

For infinitely many solutions, we must have

\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

⇒ \(\frac{5}{10}=\frac{2}{4}=\frac{-k}{-3}\)

⇒ \(\frac{1}{2}=\frac{1}{2}=\frac{k}{3}\)

⇒ \(\frac{1}{2}=\frac{k}{3}\)

⇒ k = \(\frac{3}{2}\)

Question 13.

The value of \(\frac{4}{\cot ^2 30^{\circ}}+\frac{1}{\sin ^2 60^{\circ}}\) – cos^{2} 45° is

(a) \(\frac{29}{6}\)

(b) \(\frac{77}{6}\)

(c) \(\frac{13}{6}\)

(d) \(\frac{29}{12}\)

Answer:

(c) \(\frac{13}{6}\)

\(\frac{4}{\cot ^2 30^{\circ}}+\frac{1}{\sin ^2 60^{\circ}}\) – cos^{2} 45° = \(\)

= \(\frac{4}{(\sqrt{3})^2}+\frac{1}{\left(\frac{\sqrt{3}}{2}\right)^2}-\left(\frac{1}{\sqrt{2}}\right)^2\)

= \(\frac{4}{3}+\frac{4}{3}-\frac{1}{2}\)

= \(\frac{8}{3}-\frac{1}{2}\)

= \(\frac{16-3}{6}=\frac{13}{6}\)

Question 14.

If the centre of a circle is (4, – 1) and one end of the diameter is (2, 0), then the coordinates of the other end is

(a) (10,- 3)

(b) (6, – 2)

(c) (2, – 1)

(d) (- 6, 1)

Answer:

(b) (6, – 2)

Let the coordinates of other end of a diameter be (x_{1}, y_{1})

We know that mid-point of the end of a diameter is equal to the centre of a circle.

∴ \(\left(\frac{2+x_1}{2}, \frac{0+y_1}{2}\right)\) = (4, – 1)

\(\left(\frac{2+x_1}{2}, \frac{y_1}{2}\right)\) = (4, – 1)

On equating the coordinates, we get

\(\frac{2+x_1}{2}\) = 4

and \(\frac{y_1}{2}\) = – 1

⇒ x_{1} = 8 – 2

and y = – 2

⇒ x_{1} = 6

and y_{1} = – 2

Question 15.

The zeroes of the polynomial x^{2} – √2x – 12 are

(a) √2, – √2

(b) 3√2, – 2√2

(c) – 3√2, 2√2

(d) 3√2, 2√2

Answer:

(b) 3√2, – 2√2

x^{2} – √2x – 12 = 0

⇒ x^{2} – 3√2 x + 2√2x – 12 = 0

⇒ x (x – 3√2) + 2√2 (x – 3√2) = 0

⇒ (x + 2√2) (x – 3√2) = 0

∴ x = – 2√2

or x = 3√2.

Question 16.

If α and β are zeroes of the polynomial p(x) = 4x^{2} + 3x + 7, then the value of \(\frac{1}{\alpha}+\frac{1}{\beta}\) is

(a) – \(\frac{7}{3}\)

(b) \(\frac{7}{3}\)

(c) \(-\frac{3}{7}\)

(d) \(-\frac{3}{4}\)

Answer:

(c) \(-\frac{3}{7}\)

We have, α + β = – \(\frac{3}{4}\)

and αβ = \(\frac{7}{4}\)

∴ \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{\alpha+\beta}{\alpha \beta}\)

= \(\frac{(-3 / 4)}{(7 / 4)}\)

= \(-\frac{3}{7}\)

Question 17.

If a = 2 cosec^{2} θ – 1 and b = cot^{2} θ – 3, then a – 2 b is equal to

(a) 6

(b) 7

(c) – 7

(d) 5

Answer:

(b) 7

We have,

a = 2cosec^{2} – 1

and b = cot^{2} – 3

∴ a – 2b = (2 cosec^{2} θ – 1) – 2 (cot^{2} θ – 3)

= 2(cosec^{2} θ – cot^{2} θ) – 1 + 6

= 2(1) – 1 + 6

[∵ cosec^{2} A – cot^{2}A = 1]

= 2 – 1 + 6 = 7

Question 18.

A month is selected at random in a year, the probability that it is March or October is

(a) \(\frac{1}{3}\)

(b) \(\frac{1}{4}\)

(c) \(\frac{1}{6}\)

(d) \(\frac{1}{2}\)

Answer:

(c) \(\frac{1}{6}\)

There are 12 months in a year.

∴ Required probability = \(\frac{2}{12}=\frac{1}{6}\)

Directions : In the question number 19 and 20, a statement of Assertion (A) is followed by a statement of Reason (R). Choose the correct option.

Question 19.

Assertion (A) : If x = 3 sin^{2} θ and y = 3 cos^{2} θ + 1 then the value of x + y = 4.

Reason (R) : For any value of θ, sin^{2} θ + cos^{2} θ = 1

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Given, x = 3 sin^{2} θ

and y = 3 cos^{2} θ + 1

∴ x + y = 3 sin^{2} θ + 3 cos^{2} θ + 1

= 3(sin^{2} θ + cos^{2} θ) + 1

= 3 + 1 = 4

[∵ sin^{2} A + cos^{2} A = 1]

So, the given Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Question 20.

Assertion (A) : The point (0, 7) lies on Y-axis.

Reason (R) : The x – coordinate of the point on Y-axis is zero.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

(b) Both Assertion (A) and Reason (R) are true but Reason (R) is not the correct explanation of Assertion (A).

(c) Assertion (A) is true but Reason (R) is false.

(d) Assertion (A) is false but Reason (R) is true.

Answer:

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

The x-coordinate of the given point is zero

and y – coordinate = 7

Since, the x-coordinate ¡s zero, so the point lies on Y-axis.

So, the given Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of Assertion (A).

Section – B

Section B consists of 5 questions of 2 marks each

Question 21.

A wheel has diameter 84 cm. Find how many complete revolutions must it take to cover 792 m.

Answer:

Let r be the radius of the wheel.

Then, diameter 84 cm

⇒ 2r = 84 cm

⇒ r = 42 cm

∴ Circumference of the wheel = 2πr

= 2 × \(\frac{22}{7}\) × 42

= 264 cm

= 2.64 m [∵ 1 m = 100 cm]

So, the wheel covers 264 m in one complete revolution.

∴ Total number of revolutions ¡n covering 792 m = \(\frac{792}{2.64}\) = 300.

Question 22.

A letter is chosen at random from the letters of the word ACCEPTABILITIES. Find the probability that the letter chosen is a

(i) vowel

(ii) consonant.

Answer:

There are 15 letters in the word ACCEPTABILITIES out of which one letter can be chosen in 15 ways.

∴ Total number of elementary events = 15

(i) There are 7 vowels in the word ACCEPTABILITIES.

So, there are 7 ways of selecting a vowel.

∴ Probability of selecting a vowel = \(\frac{7}{15}\)

(ii) We have,

Probability of selecting a consonant = 1 – Probability of selecting a vowel

= 1 – \(\frac{7}{15}\)

= \(\frac{8}{15}\)

Or

Ashok buys a fish from a shop for his aquarium. The shopkeeper takes out one fish at random from a tank containing 10 male ‘ fish and 9 female fish. What is the probability that the fish taken out is a male fish?

Answer:

There ace 19 = (10 + 9) fish out of which one can be chosen in 19 ways.

∴. Total number of elementary events = 19

There are 10 maIe fish out of which one male fish can be chosen in 10 ways.

∴ Favourable number of elementary events = 10

Hence, required probability = \(\frac{10}{19}\)

Question 23.

The mid-point P of the line segment joining the points A (- 10, 4) and B (- 2, 0) lies on the line segment joining the points C (- 9, – 4) and D (- 4, y). Find the ratio in which P divides CD. Also, find the value of y.

Answer:

Given the mid-point P of the line segment joining the points A (- 10, 4) and B (- 2, 0).

The mid-point of P = \(\left(\frac{-10-2}{2}, \frac{4+0}{2}\right)\) = (-6, 2)

Let P (- 6 2) divide CD in the ratio k : 1.

Thus, \(\frac{-4 k-9}{k+1}\) = – 6

and \(\frac{k y-4}{k+1}\) = 2

⇒ – 4k – 9 = – 6k – 6

and ky – 4 = 2k + 2

⇒ 2k = 3

and k (y – 2) = 6

⇒ k = \(\frac{3}{2}\)

and (y – 2) = \(\frac{6}{k}\)

= 6 × \(\frac{2}{3}\) = 4

k = \(\frac{3}{2}\)

and y = 4 + 2 = 6

Hence, the ratio is 3 : 2 and y = 6.

Question 24.

If the mean of x and \(\frac{1}{x}\) is M, then what is the value of mean of x^{3} and \(\frac{1}{x^3}\)?

Answer:

Given, mean of x and \(\frac{1}{x}\) is M.

Question 25.

Find the solution of the equation \(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}\), x ≠ 3, 4.

Answer:

Given, equation is \(\frac{2 x}{x-4}+\frac{2 x-5}{x-3}=\frac{25}{3}\), x ≠ 3, 4

⇒ \(\frac{2 x(x-3)+(x-4)(2 x-5)}{(x-4)(x-3)}=\frac{25}{3}\)

⇒ \(\frac{2 x^2-6 x+\left(2 x^2-13 x+20\right)}{(x-4)(x-3)}=\frac{25}{3}\)

⇒ 3(4x^{2} – 19x + 20) = 25 (x^{2} – 7x + 12)

⇒ 25x^{2} – 12x^{2} – 175x + 57x + 300 – 60 = 0

⇒ 13x^{2} – 118x + 240 = 0

⇒ 13x^{2} – 78x – 40x + 240 = 0 [by factorisation method]

⇒ 13x (x – 6) – 40 (x – 6) = 0

⇒ (13x – 40) (x – 6) = 0

⇒ 13x – 40 = 0 or x – 6 = 0

⇒ x = \(\frac{40}{13}\) or x = 6

Or

If \(\frac{1}{2}\) is a root of the equation x^{2} + kx – \(\frac{5}{4}\) = 0, then find the value of k.

Answer:

Since, \(\frac{1}{2}\) is a root of the quadratic equation.

x^{2} + kx – \(\frac{5}{4}\) = 0

∴ \(\left(\frac{1}{2}\right)^2+k\left(\frac{1}{2}\right)-\frac{5}{4}\) = 0

\(\frac{1}{4}+\frac{k}{2}-\frac{5}{4}\) = 0

⇒ \(\frac{1+2 k-5}{4}\) = 0

⇒ 2k – 4 = 0

⇒ 2k = 4

∴ k = 2.

Section – C

Section C consists of 6 questions of 3 marks each

Question 26.

If α and β are the zeroes of the quadratic polynomial f(x) = kx^{2} + 4x + 4 such that 2 α^{2} + β^{2} = 24, then find the values of k.

Answer:

Since, α and β is a root of the quadratic equation.

polynomial f(x) = kx^{2} + 4x + 4

∴ α + β = – \(\frac{4}{k}\)

and αβ = \(\frac{4}{k}\)

Now, given

α^{2} + β^{2} = 24

⇒ (α + β)^{2} – 2αβ = 24

⇒ \(\left(-\frac{4}{k}\right)^2-2 \times \frac{4}{k}\) = 24

⇒ \(\frac{16}{k^2}-\frac{8}{k}\) = 24

⇒ 16 – 8k = 24k^{2}

⇒ 3k^{2} + k – 2 = 0

⇒ 3k^{2} + 3k – k – 2 = 0

⇒ 3k (k + 1) – 2 (k + 1) = 0

⇒ (k + 1) (3k – 2) = 0

⇒ k + 1 = 0 or 3k – 2 = 0

⇒ k = – 1 or k = \(\frac{2}{3}\)

Hence, k = – 1 or k = \(\frac{2}{3}\).

Question 27.

Prove that √2 + √5 is irrational, given that √2 is irrational.

Answer:

Let us assume on the contrary that √2 + √5 be a rational number.

Then, there exist coprime positive integers a and b such that

√2 + √5 = \(\frac{a}{b}\)

⇒ \(\frac{a}{b}\) – √2 = √5

⇒ \(\left(\frac{a}{b}-\sqrt{2}\right)^2\) = (√5)^{2} [squaring both sides]

⇒ \(\frac{a^2}{b^2}-\frac{2 a}{b}\) √2 + 2 = 5

⇒ \(\frac{a^2}{b^2}-3=\frac{2 a}{b} \sqrt{2}\)

⇒ \(\frac{a^2-3 b^2}{2 a b}\) = √2

√2 is a rational number.

[∵ a, b are integers

∵\(\frac{a^2-3 b^2}{2 a b}\) is rational]

This contradicts the fact that √2 is irrational.

So, our assumption is wrong.

Hence, √2 + √5 is irrational.

Hence proved.

Question 28.

Find the value of \(\frac{\tan ^2 60^{\circ}+4 \cos ^2 45^{\circ}+3 \sec ^2 30^{\circ}+5 \cos ^2 90^{\circ}}{\ {cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^2 30^{\circ}}\).

Answer:

\(\frac{\tan ^2 60^{\circ}+4 \cos ^2 45^{\circ}+3 \sec ^2 30^{\circ}+5 \cos ^2 90^{\circ}}{\ {cosec} 30^{\circ}+\sec 60^{\circ}-\cot ^2 30^{\circ}}\)

= \(\frac{(\sqrt{3})^2+4\left(\frac{1}{\sqrt{2}}\right)^2+3\left(\frac{2}{\sqrt{3}}\right)^2+5(0)}{2+2-(\sqrt{3})^2}\)

= \(\frac{3+\frac{4}{2}+3 \times \frac{4}{3}+0}{4-3}\)

= \(\frac{3+2+4}{1}\)

= 9

Or

If cos(A + B) = 0 and cot (A – B) = √3, then find the value of cos A . cos B – sin A . sin B

Answer:

Given, cos (A + B) = 0

⇒ cos (A + B) = cos 90°

⇒ A + B = 90° ………………(i)

and cot (A – B) =

⇒ cot (A – B) = cot 30°

⇒ A – B = 30° ………………(ii)

On solving Eqs. (i) and (ii), we get

∴ A = 60°

and B = 30°

∴ cos A . cos B – sin A . sin B = cos 60° . cos 30° – sin 60° sin 30°

= \(\frac{1}{2} \times \frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2} \times \frac{1}{2}\)

= \(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\)

= 0

Question 29.

One number of two digits is such that the product of digits is 12. If 36 is added to the number, then the places of the digits are interchanged. Find the number.

Answer:

Let the units place digit be y and tens place digit be x.

Then, number of two digits = 10x + y

∵ Product of digits. xy = 12 [given] …………….(i)

Interchanging the position of digits, new number = 10y + x

According to the question,

10x + y + 36 = 10y + x

⇒ 10x – x + y- 10y + 36 = 0 [on transposition]

⇒ 9x – 9y + 36 = 0

⇒ x – y + 4 = 0 [on dividing by 9]

⇒ y = x + 4 …………….(ii)

On putting the value of y from Eq (ii) in Eq. (i), we get

x (x + 4) = 12

⇒ x^{2} + 4x = 12

⇒ x^{2} + 4x – 12 = 0 [on transposition]

By using factorisation method,

⇒ x^{2} + (6 – 2) x – 12 = 0

⇒ x^{2} + 6x – 2x – 12 = 0

⇒ x (x + 6) – 2 (x + 6) = 0

⇒ (x + 6) (x – 2) = 0

⇒ x + 6 = 0 or x – 2 = 0

⇒ x = – 6 or x = 2

∴ x = 2

[∵ x ≠ – 6]

On putting the value of x in Eq. (ii), we get

y = x + 4

= 2 + 4 = 6

Hence, required number of two digits = 10x + y

= 10 × 2 + 6

= 20 + 6 = 26

Question 30.

A circle is touching the side BC of ∆ABC at P and touching AB and AC produced at Q and R, respectively. Prove that AQ = \(\frac{1}{2}\) (Perimeter of ∆ABC).

Answer:

Since, tangent from an exterior point to a circle are equal in length.

∴ BP = BQ [from B] ……………(i)

CP = CR [from C] ……………..(ii)

and AQ = AR [from A] ………….(iii)

From Eq. (ii), we have

AQ = AR

⇒ AB + BQ = AC + CR

⇒ AB + BP = AC + CP [using Eqs. (i) and (ii)] ………………… (iv)

Now, Perimeter of ∆ABC= AB + BC + AC

⇒ Perimeter of ∆ABC = AB + (BP + PC) + AC

⇒ Perimeter of ∆ABC = (AB + BP) + (AC + PC)

⇒ Perimeterof ∆ABC = 2 (AB + BP) [using Eq. (iv)]

⇒ Perimeter of ∆ABC = 2 (AB + BQ) [using Eq. (i)]

⇒ Perimeter of ∆ABC = 2AQ

∴ AQ = \(\frac{1}{2}\) (Perimeter of ∆ABC)

Hence proved.

Or

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the bigger circle. BD is a tangent to the smaller circle touching it at D. Find the length AD.

Answer:

Let the line BD intersects the bigger circle at E.

Now, Join AE.

Let O be the centre of the bigger circle, then O is the mid-point of AB.

[∵ AB is a diameter of the bigger circle]

BD is a tangent to the smaller circle andOD s a radius through the point of contact D.

Then,

OD ⊥ BD

⇒ OD ⊥ BE

Since, OD is perpendicular to a chord BE of bigger circle.

∴ BD = DE

[∵ perpendicular drawn from the centre to a chord bisects the chord]

D is the mid-point of BE.

∴ In ∆BAE, O is the mid-point of AB

and D is the mid-point of BE.

∴ OD = AE

[∵ line segment joining the mid-points of any two sides of a triangle is half of the third side]

⇒ AE = 2 (OD)

= 2 × 8

= 16 cm

In right angled ∆OBD. use Pythagoras theorem,

OD^{2} + BD^{2} = OB^{2}

BD = \(\sqrt{O B^2-O D^2}\)

= \(\sqrt{(13)^2-(8)^2}\)

= \(\sqrt{169-64}\)

= \(\sqrt{105}\)

∴ DE = BD = \(\sqrt{105}\) cm

In right angled ∆AED, using Pythagoras theorem, we have

AD = \(\sqrt{A E^2+D E^2}\)

= \(\sqrt{(16)^2+(\sqrt{105})^2}\)

= \(\sqrt{256+105}\)

= \(\sqrt{361}\)

= 19 cm

Question 31.

Find the value of the height (h) in the given figure, at which the tennis ball must be hit, so that it will just pass over the net and land 7 m away from the base of the net.

Answer:

Since, the height (h) is measured vertically.

So, ∠EDA is a right angle.

We assume that the net (i.e. CB) is vertical.

Here, ∆ADE and ∆ABC are similar.

[by AA similarity criterion]

∴ \(\frac{D E}{B C}=\frac{A D}{A B}\)

\(\frac{h}{0.8}=\frac{17}{7}\)

h = \(\frac{17 \times 0.8}{7}\)

h = 1.94 m

Hence, the height at which the ball should be hit is 19.4 m.

Section – D

Section D consists of 4 questions of 5 marks each

Question 32.

CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG, respectively.

If ∆ABC ~ ∆FEG, then show that

(i) \(\frac{C D}{G H}=\frac{A C}{F G}\)

(ii) ∆ DCB ~ ∆ HGE

Answer:

Draw ∆ABC and ∆FEG and then draw bisectors CD and GH of ∠ACB and ∠EGF such that D lies on AB and H lies on FE.

(i) Given, ∆ABC ~ ∆FEG

So, all corresponding angles are equal.

∴ ∠CAB = ∠GFE

or ∠CAD = ∠GFH …………..(i)

and ∠ACB = ∠FGE

⇒ \(\frac{1}{2}\) ∠ACB = \(\frac{1}{2}\) ∠FGE

[dividing both sides by 2]

⇒ ∠ACD = ∠FGH …………… (ii)

From Eqs. (i) and (ii), we get

∆ACD ~ ∆FGH [by AA similarity criterion]

∴ \(\frac{C D}{G H}=\frac{A C}{F G}\)

[since, corresponding sides of two similar triangles are proportional]

(ii) In ∆DCB and ∆HGE,

[∠DBC = ∠HEG ……………(iii)

∴ ∆ABC = ∆FEG

⇒ ∠DBC = ∠HEG

and ∠DCB = ∠HGE] ……………. (iv)

[∵ ∆ABC ~ ∆FEG

∴ ∠ACB = ∠FGE

⇒ \(\frac{1}{2}\) ∠ACB = \(\frac{1}{2}\) ∠FGE

⇒ ∠DCB = ∠HGE

From Eqs. (iii) and (iv),

∆DCB ~ ∆HGE [by AA similarity criterion]

Or

Sides AB and BC and median AD of ∆ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR. Show that ∆ABC ~ ∆PQR.

Answer:

Given : ∆ABC and ∆PQR in which AD and PM are their medians respectively

and \(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}\)

To prove : ∆ABC ~ ∆PQR

Proof : We have,

\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A D}{P M}\) [given]

⇒ \(\frac{A B}{P Q}=\frac{1 / 2 B C}{1 / 2 Q R}=\frac{A D}{P M}\)

[multiplying numerator and denominator of middle term by 1/2]

⇒ \(\frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}\)

[since, medians AD and PM bisect the lines BC and OR, respectively

i.e. BD = 1/2 BC and QM = 1/2 QR]

⇒ ∆ADB ~ ∆PMQ (by SSS similarity criterion]

∴ ∠B = ∠Q ………………(i)

In ∆ABC and ∆PQR,

\(\frac{A B}{P Q}=\frac{B C}{Q R}\) [given]

and ∠B = ∠Q [from Eq. (i)]

∴ ∆ABC ~ ∆PQR

[by SAS similarity criterion]

Hence Proved.

Question 33.

A toy is in the form of a cone mounted on a hemisphere of radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer:

Let r be the radius of cone and hemisphere and H be the height of the toy.

Given, radius of the conical and hemispherical portion,

r = 3.5 cm

and total height of the toy, H = 15.5 cm

∴ Height of conical portion,

h = Total height – Radius of hemispherical portion

= 15.5 – 3.5 = 12 cm

∴ Slant height of the cone,

l = \(\sqrt{h^2+r^2}\)

= \(\sqrt{(12)^2+(3.5)^2}\)

= \(\sqrt{144+12.25}\)

= \(\sqrt{156.25}\)

l = 12.5 cm

Now, total surface area of the toy = Curved surface area of conical portion + Curved surface area of hemispherical portion

= πrl + 2πr^{2}

= \(\frac{22}{7}\) × 3.5 × (12.5 + 2 × 3.5)

= \(\frac{22}{7}\) × 3.5 × 19.5

= 214.5 cm^{2}

Hence, the total surface area of the toy is 214.5 cm^{2} .

Question 34.

During the medical check up of 35 students of a class, their weight were recorded as follows.

Weight (in kg) | Number of students |

Less than 38 | 0 |

Less than 40 | 3 |

Less than 42 | 5 |

Less than 44 | 9 |

Less than 46 | 14 |

Less than 48 | 28 |

Less than 50 | 32 |

Less than 52 | 35 |

Answer:

Here, we have cumulative frequency distribution less than types.

First we convert it into an ordinary frequency distribution, which is shown below.

Weight (in kg) | Frequency | Cumulative frequency |

38 – 40 | 3 | 3 |

40 – 42 | 5 – 3 = 2 | 5 |

42 – 44 | 9 – 5 = 4 | 9 |

44 – 46 | 14 9 = 5 | 14(cf) |

46 – 48 | 28 – 14 = 14(f) | 28 |

48 – 50 | 32 – 28 = 4 | 32 |

50 – 52 | 35 – 32 = 3 | 35 |

Total | N = 35 |

Here, \(\frac{N}{2}=\frac{35}{2}\) = 17.5

Since, the cumulative frequency just greater than 17.5 is 28 and the corresponding class-interval is 46 – 48.

∴ Median class is 46 – 48

So, l = 46,

\(\frac{N}{2}\) = 17.5,

f = 14

and cf = 14

Now, median = l + \(\frac{\left(\frac{N}{2}-c f\right)}{f}\) × h

= 46 + \(\frac{17.5-14}{14}\) × 2

= 46 + \(\frac{3.5}{7}\)

= 46 + \(\frac{1}{2}\)

=46 + 0.5 = 46.5 kg

Or

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows

Find the median of the above data.

Answer:

The cumulative frequency table of given data is

Number of letters | Number of surnames (f_{i}) |
Cumulative frequency (cf) |

0 – 5 | 10 | 10 |

5 – 10 | 30 | 40(cf) |

10 – 15 | 40 (f) | 80 |

15 – 20 | 16 | 96 |

20 – 25 | 4 | 100 |

Total | N = 100 |

Here, N =100

∴ \(\frac{N}{2}=\frac{100}{2}\) = 50

Since, the cumulative frequency just greater than 50 is 80 and the corresponding class interval is 10 – 15.

∴ Median class is 10 – 15

∴ l = 10

cf = 40,

h = 5

and f = 40

Now, median = l + \(\left(\frac{\frac{N}{2}-c f}{f}\right)\) × h

= 10 + \(\left(\frac{50-40}{40}\right)\) × 5

= 10 + \(\frac{10}{40}\) × 5

= 10 + \(\frac{5}{4}\)

= 10 + 1.25 = 11.25

Question 35.

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the cable tower.

Answer:

Let AD = 7 m be the height of the building and

BC = h m be the height of the cable tower.

From the top of the building D, the angles of elevation

and depression are ∠CDE = 60° and ∠EDB = 45°.

From the point D, draw a line DE II AB.

Then, ∠EDB = ∠ABD = 45° [alternate angles]

In right angled ∆BAD,

tan 45° = \(\frac{AD}{AB}\)

[∵ tan θ = Perpendicular / Base]

⇒ 1 = \(\frac{7}{x}\)

[∵ tan 45° = 1]

⇒ x = 7m ……………(i)

Now, in right angled ∆CED,

tan 60° = \(\frac{C E}{D E}\)

= \(\frac{C B-B E}{A B}\)

⇒ √3 = \(\frac{h-7}{x}\)

[∵ tan 60° = √3]

⇒ h – 7 = x√3

⇒ h = x√3 + 7

⇒ h = 7√3 +7 [from Eq. (i)]

⇒ h = 7(√3 + 1) m

Hence, the height of the cable tower is 7 (√3 + 1) m.

Section – E

Case study based questions are compulsory

Question 36.

Sonu is studying in X standard. He is making a brooch and understanding areas related to circles. A brooch is made with silver wire in the form of a circle with diameter 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors.

Few questions came in his mind while making the brooch. Give answers to his questions.

(i) Find the area of circle.

Answer:

∴ Area of circle = πr²

= π \(\)

= \(\)

= 11 × 5 × \(\frac{35}{2}\)

= \(\frac{1925}{2}\)

= 962.5 mm^{2}

(ii) Find the measure of angle of each sector.

Answer:

Since, total circle is divided into 10 sectors

∴ Angle of each sector = \(\frac{360^{\circ}}{10}\) = 36°

(iii) Find the area of each sector of the brooch.

Answer:

(iii) Area of each sector of the brooch = \(\frac{\theta}{360^{\circ}}\) × πr²

= \(\frac{36^{\circ}}{360^{\circ}} \times \frac{22}{7}\left(\frac{35}{2}\right)^2\)

[∵ r = \(\frac{d}{2}\)

= \(\frac{35}{2}\) mm]

= \(\frac{1}{10} \times \frac{22}{1} \times \frac{5}{2} \times \frac{35}{2}\)

= \(\frac{11 \times 35}{2 \times 2}\)

= \(\frac{385}{4}\) mm^{2}

Or

Find the total length of the silver wire.

Answer:

Circumference of circle = πd

= \(\frac{22}{7}\) × 35

= 110 mm

Total length of the silver wire = πd + 5d

= 110 + 5 × 35

= 285 mm

Question 37.

Ram asks the labour to dig a well up to a depth of 15 m. Labour charges ? 100 for first metre and increase ₹ 25 for each subsequent metres. As labour was uneducated, he claims ₹ 400 for the whole work.

On the basis of above information, answer the following questions.

(i) Find the AP sequence that is formed by the labour charge.

Answer:

For first metre. the charge is ₹ 100

i.e. first term, a = 100

As, there is increasing of ₹ 25 for each subsequent

metres, therefore common difference, d = 25

So, the AP thus formed is 100, 125, 150, ………………

(ii) What is the third term of the AP sequence thus formed.

Answer:

The third term of the AP sequence is 150

Or

Find the sum of AP sequence that is formed by the labour charge.

Answer:

We know that

S_{n} = \(\frac{n}{2}\) [2a + (n – 1) d]

Sum of 15 terms,

S_{15} = \(\frac{15}{2}\) [2 × 100 + 14 × 25]

= \(\frac{15}{2}\) [200 + 350]

= \(\frac{15}{2}\) × 550

= 4125

(iii) Find the labour charge to dig the well.

Answer:

Labour charge to dig the well is the 15th term of AP

We know, a_{n} = a + (n – 1) d

∴ a_{15} = 100 + (15 – 1) × 25

= 100 + 14 × 25 = 450

∴ Labour charge = ₹ 450

Question 38.

It is common that Government revise travel fares from time to time based on various factors such as inflation (a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto rickshaws, taxis, radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered.

Study the following situations:

Situation 1

In City A, for a journey of 10 km, the charge paid is ₹ 75 and for a journey of 15 km, the charge paid is ₹ 110.

Situation 2

In City B, for a journey of 8 km, the charge paid is ₹ 91 and for a journey of 14 km, the charge paid is ₹ 145.

On the basis of above information, answer the following questions.

**Refer Situation**

(i) If the fixed charges of auto rickshaws is ₹ x and the running charges be ₹ y per km, then find the pair of linear equations representing this situation.

Answer:

Fixed charges = ₹ x

and running charges = ₹ y per km

∴ Running charges for 10 km = 10y

and for 15 km = 15y

∴ Total amount paid = Fixed charges + Running charges

⇒ x + 10y = 75 …………(i)

and x + 15y = 110 …………(ii)

(ii) A person travels a distance of 50 km. Find the amount he has to pay.

Answer:

On subtracting Eqs. (i) and (ii), we get

5y = 35

⇒ y = 7

On substituting y = 7 in Eq. (i), we get

x + 70 = 75

⇒ x = 5

∴ Fixed charges = ₹ 5

and running charges = ₹ 7 per km

So, for 50 km,

x + 50y = 5 + 50 × 7

= 5 + 350

= 355

∴ Amount he has to pay is ₹ 355.

**Refer Situation 2**

(iii) Draw the graph of lines representing the conditions (Situation 2).

Answer:

Let fixed ctiarges = ₹ x

and running charges = ₹ y per km

Running charges for 8 km = 8y

and for 14 km = 14y

So, the equations are

x + 8y = 91 …………..(i)

and x + 14y = 145 …………..(ii)

Taking Eq. (i)

Now, taking Eq. (i),

Or

Solve the pair of linear equations formed (Situation 2) by elimination method.

Answer:

Pair of linear equations representing the situation are

⇒ y = 9

On putting y = 9 in Eq. (i), we get

x + 8 (9) = 91

⇒ x = 91 – 72 = 19

Hence, x = 19 and y = 9.